Let's solve each part step by step: --- ## (a) Where is \( F(x) = \int_{-4}^{x} f(t)dt \) **decreasing** on \([-4,2]\)? Recall: \( F(x) \) is **decreasing** where \( F'(x) = f(x) < 0 \). So, we look for intervals where the graph of \( f(x) \) is **below the x-axis**. **From the graph:** - \( f(x) < 0 \) roughly on the intervals \( (-3.6, -2.2) \), \( (-0.6, 0.6) \), and \( (1.7, 2) \). So, the answer is: \[ \boxed{B. (-3.6, -2.2),\ (-0.6, 0.6),\ (1.7, 2)} \] --- ## (b) Estimate \( \int_{-1}^{1} f(x)\, dx \): Look at the area between \( x = -1 \) and \( x = 1 \). - The positive and negative areas look roughly equal in size, but the positive area (left of \( x = 0 \)) seems a bit larger. - Thus, the total signed area will be slightly **greater than 0**, but less than 1. From the options, the closest value is: \[ \boxed{C. \ 0.5} \] --- ## (c) Given \( F(-1) = -5 \), estimate \( \int_{6}^{11} f(7-x) dx \): Let’s rewrite the integral by **substitution**: Let \( u = 7-x \implies du = -dx \). - When \( x = 6 \), \( u = 1 \). - When \( x = 11 \), \( u = -4 \). So, \[ \int_{6}^{11} f(7-x) dx = \int_{x=6}^{x=11} f(7-x) dx = \int_{u=1}^{u=-4} f(u) (-du) = \int_{u=-4}^{u=1} f(u) du \] So, \[ \int_{6}^{11} f(7-x) dx = \int_{-4}^{1} f(u) du = F(1) \] But, \[ F(-1) = \int_{-4}^{-1} f(t)dt = -5 \] \[ F(1) = \int_{-4}^{1} f(t)dt = \int_{-4}^{-1} f(t)dt + \int_{-1}^1 f(t)dt = F(-1) + \int_{-1}^1 f(t)dt \] From (b), \(\int_{-1}^1 f(t)dt \approx 0.5\). So, \[ F(1) = -5 + 0.5 = -4.5 \] Thus, \[ \boxed{-4.5} \] --- ## **Final Answers** (a) **B**. \((-3.6, -2.2),\ (-0.6, 0.6),\ (1.7, 2)\) (b) **C**. \(0.5\) (c) \(\boxed{-4.5}\)

Let's go through each part carefully, explaining every step and calculation. --- ## **Part (a): Determining where \( F(x) = \int_{-4}^{x} f(t) dt \) is decreasing on \([-4, 2]\)** ### Concept: - The function \( F(x) \) is decreasing where its derivative \( F'(x) = f(x) \) is **less than zero**. - Therefore, **\( F(x) \) decreases where \( f(x) < 0 \)**. ### Step 1: Analyze the graph of \( f(x) \) From the graph (based on the image), approximate the regions where \( f(x) \) is negative: - **Interval 1**: roughly from \( x \approx -3.6 \) to \( x \approx -2.2 \) - **Interval 2**: roughly from \( x \approx -0.6 \) to \( x \approx 0.6 \) - **Interval 3**: roughly from \( x \approx 1.7 \) to \( x = 2 \) These are intervals where \( f(x) < 0 \), so \( F(x) \) is decreasing in these regions. ### **Answer:** \[ \boxed{\textbf{Option B:}\ (-3.6, -2.2),\ (-0.6, 0.6),\ (1.7, 2)} \] --- ## **Part (b): Estimating \( \int_{-1}^{1} f(x) dx \)** ### Concept: - The integral represents the **signed area** between the graph of \( f(x) \) and the x-axis over \([-1, 1]\). - From the graph, estimate the area: ### Step 2: Approximate the areas - From \( x = -1 \) to \( x = 0 \): The graph is **above** the x-axis, roughly a triangle or trapezoid with a height of about 1 unit, width 1. So area ≈ 0.5. - From \( x = 0 \) to \( x = 1 \): The graph is **below** the x-axis, roughly similar in magnitude, but negative. Area ≈ -0.4 to -0.5. - Total approximate integral: \[ \text{Area} \approx 0.5 + (-0.5) \approx 0 \] But considering the graph's shape, the positive area is slightly larger, so the total is **around 0.5**. ### **Answer:** \[ \boxed{\textbf{Option C:}\ 0.5} \] --- ## **Part (c): Computing \( \int_{6}^{11} f(7 - x) dx \)** ### Step 1: Substitution to simplify the integral Let: \[ u = 7 - x \implies du = -dx \] Change the limits: - When \( x = 6 \): \[ u = 7 - 6 = 1 \] - When \( x = 11 \): \[ u = 7 - 11 = -4 \] Now, rewrite the integral: \[ \int_{6}^{11} f(7 - x) dx = \int_{x=6}^{x=11} f(7 - x) dx \] In terms of \( u \): \[ = \int_{u=1}^{u=-4} f(u) \cdot (-du) = \int_{u=-4}^{u=1} f(u) du \] ### Step 2: Recognize the integral \[ \int_{-4}^{1} f(u) du = F(1) - F(-4) \] but \( F(x) = \int_{-4}^{x} f(t) dt \). So, \[ \int_{-4}^{1} f(u) du = F(1) \] since \( F(-4) = 0 \) (by the definition of the indefinite integral starting at \(-4\)). ### Step 3: Find \( F(1) \) Given: \[ F(-1) = \int_{-4}^{-1} f(t) dt = -5 \] From part (b), the estimate for \( \int_{-1}^{1} f(t) dt \) is approximately 0.5. Note: \[ F(1) = \int_{-4}^{1} f(t) dt = \int_{-4}^{-1} f(t) dt + \int_{-1}^{1} f(t) dt = F(-1) + \text{area from } -1 \text{ to } 1 \] \[ F(1) = -5 + 0.5 = -4.5 \] ### **Final calculation:** \[ \int_{6}^{11} f(7 - x) dx = F(1) = -4.5 \] --- ## **Summary of solutions:** | Part | Final Answer | Explanation | |--------|----------------|-------------| | (a) | **B** | \( f(x) < 0 \) on the given intervals, so \( F(x) \) decreases there. | | (b) | **C** | The area between \( -1 \) and \( 1 \) is about 0.5. | | (c) | **-4.5** | Using substitution, the integral equals \( F(1) \), estimated as \( -4.5 \). | --- **Hope this detailed explanation helps!**